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Line L in the xy-plane contains points A and B with coordinates (-4,5) and (6,-1), respectively. Line k is perpendicular to L and contains the midpoint of line segment AB.

Which of the following statements are true?

I. The slope of line l is -3/5.
II. Line k has a negative slope.
III. Line k contains the point (1,2).

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

Slope of L = y2-y1/ x2-x1 = (-1-5) / [6-(-4) ]
Slope L = -3/5
St. 1 is correct

Slope of K= Negative reciprocal of slope of L (Since perpendicular lines)
Slope K = 5/3
St. 2 is incorrect

To see if point (1,2) lies on K, we need to set up the linear equation for K
Start with the only known point for K, i.e. Mid-point of AB

Mid-point AB = Mid-point x-axis, Mid-point y-axis
Distance of mid-point of AB= -4-6/ 2 , 5-(-1)/2 = -5, 3

Since this is the distance, to locate the point subtract the distance from either point A or point B
=> -4-(-5), 5- 3 = [1, 2]

St. 3 is correct.

Answer is option D.


If we subtract the distance, from point (6,-1), we will get (6-(-5),-1-3) = (11,-4).
kindly help me understand.
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Can someone please help with this as I didn't understand the following statements:
Mid-point AB = Mid-point x-axis, Mid-point y-axis
Distance of mid-point of AB= -4-6/ 2 , 5-(-1)/2 = -5, 3
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arorni
Can someone please help with this as I didn't understand the following statements:
Mid-point AB = Mid-point x-axis, Mid-point y-axis
Distance of mid-point of AB= -4-6/ 2 , 5-(-1)/2 = -5, 3

The mid point is essential to the solution and this method need not be taken at all when you have a simpler solution to find out the midpoint of AB.
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arorni
Can someone please help with this as I didn't understand the following statements:
Mid-point AB = Mid-point x-axis, Mid-point y-axis
Distance of mid-point of AB= -4-6/ 2 , 5-(-1)/2 = -5, 3

Hello, pratik2018, I don't understand this either but I came up with easier solution:
there is one good formula for finding coordination of mid-point:

x-axis would be: \(\frac{(x1+x2)}{2}\)
y-axis would be: \(\frac{(y1+y2)}{2}\)

So, mid-point of AB: (\(\frac{(-4+6)}{2}\); \(\frac{(5-1)}{2}\)) -> (1;2)
Since statement III says "Line k contains the point (1,2)" we can say that III is true.
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